nathaniel e. helwig - umn statisticsusers.stat.umn.edu/~helwig/notes/aov2-notes.pdf ·...

Factorial and Unbalanced Analysis of Variance

Nathaniel E. Helwig

Assistant Professor of Psychology and StatisticsUniversity of Minnesota (Twin Cities)

Updated 04-Jan-2017

Nathaniel E. Helwig (U of Minnesota) Factorial & Unbalanced Analysis of Variance Updated 04-Jan-2017 : Slide 1

Copyright

Copyright © 2017 by Nathaniel E. Helwig


Outline of Notes

1) Balanced Two-Way ANOVA:Model Form & AssumptionsLeast-Squares EstimationBasic InferenceHypertension Example (pt 1)Multiple ComparisonsHypertension Example (pt 2)

2) Balanced Three-Way ANOVA:Model Form & EstimationHypertension Example (pt 3)

3) Unbalanced ANOVA Models:Overview of problemTypes of sums-of-squaresHypertension Example (pt 4)


Balanced Two-Way ANOVA

Balanced Two-Way ANOVA


Balanced Two-Way ANOVA Model Form & Assumptions

Two-Way ANOVA Model (cell means form)

The Two-Way Analysis of Variance (ANOVA) model has the form

yijk = µjk + eijk

for i ∈ {1, . . . ,njk}, j ∈ {1, . . . ,a}, and k ∈ {1, . . . ,b} whereyijk ∈ R is real-valued response for i-th subject in factor cell (j , k)

µjk ∈ R is real-valued population mean for factor cell (j , k)

eijkiid∼ N(0, σ2) is a Gaussian error term

njk is number of subjects in cell (j , k) and n =∑a

j=1∑b

k=1 njk(note: njk = n∗∀j , k in balanced two-way ANOVA)a and b are number of levels for first and second factors

Implies that yijkind∼ N(µjk , σ

2).



Two-Way ANOVA Model (effect coding)

Using effect coding, the mean for factor cell (j , k) has the form

µjk = µ+ αj + βk + (αβ)jk

for j ∈ {1, . . . ,a} and k ∈ {1, . . . ,b} whereµ is overall population meanαj is main effect of first factor such that

∑aj=1 αj = 0

βk is main effect of second factor such that∑b

k=1 βk = 0(αβ)jk is interaction effect such that

∑aj=1(αβ)jk = 0 ∀k and∑b

k=1(αβ)jk = 0 ∀j

Set (αβ)jk = 0 ∀j , k to fit an additive model.



Two-Way ANOVA Model (matrix form)In matrix form, the two-way ANOVA model is y = Xb + e where

X =

1 x11 · · · x1(a−1) z11 · · · z1(b−1) x11z11 · · · x1(a−1)z11 · · · x11z1(b−1) · · · x1(a−1)z1(b−1)1 x21 · · · x2(a−1) z21 · · · z2(b−1) x21z21 · · · x2(a−1)z21 · · · x21z2(b−1) · · · x2(a−1)z2(b−1)1 x31 · · · x3(a−1) z31 · · · z3(b−1) x31z31 · · · x3(a−1)z31 · · · x31z3(b−1) · · · x3(a−1)z3(b−1)

.

.

....

. . ....

.

.

.. . .

.

.

....

. . ....

. . ....

. . ....

1 xn1 · · · xn(a−1) zn1 · · · zn(b−1) xn1zn1 · · · xn(a−1)zn1 · · · xn1zn(b−1) · · · xn(a−1)zn(b−1)

b =

(µ α1 · · · αa−1 β1 · · · βb−1 (αβ)11 · · · (αβ)(a−1)1 · · · (αβ)1(b−1) · · · (αβ)(a−1)(b−1)

)′where X has 1 + (a− 1) + (b − 1) + (a− 1)(b − 1) = ab columns

xij =

1 if i-th observation is in j-th level of first factor−1 if i-th observation is in a-th level of first factor

0 otherwise

zik =

1 if i-th observation is in k -th level of second factor−1 if i-th observation is in b-th level of second factor

0 otherwisei ∈ {1, . . . ,n} and additional subscripts on y and e are dropped

Implies that y ∼ N(Xb, σ2In).Nathaniel E. Helwig (U of Minnesota) Factorial & Unbalanced Analysis of Variance Updated 04-Jan-2017 : Slide 7


Two-Way ANOVA Model (assumptions)

The fundamental assumptions of the two-way ANOVA model are:1 xij , zik and yi are observed random variables (known constants)

2 eiiid∼ N(0, σ2) is an unobserved random variable

3 µjk are unknown constants

4 (yi |xij , zik )ind∼ N(µjk , σ

2)note: homogeneity of variance

Interpretation of µjk depends on model formAdditive: µjk = µ+ αj + βk

Interaction: µjk = µ+ αj + βk + (αβ)jk


Balanced Two-Way ANOVA Least-Squares Estimation

Ordinary Least-Squares

We want to find the effect estimates (i.e., µ, αj , βk , and (αβ)jk terms)that minimize the ordinary least squares criterion

SSE =b∑

k=1

a∑j=1

njk∑i=1

(yijk − µ− αj − βk − (αβ)jk )2

If njk = n∗∀j , k the least-squares estimates have the form

µ =1

abn∗∑b

k=1

∑aj=1

∑n∗i=1 yijk = y··

αj =

(1

bn∗∑b

k=1

∑n∗i=1 yijk

)− µ = yj· − y··

βk =

(1

an∗∑a

j=1

∑n∗i=1 yijk

)− µ = y·k − y··

(αβ)jk =

(1n∗∑n∗

i=1 yijk

)− µ− αj − βk = yjk − yj· − y·k + y··

which implies that yijk = yjk for all (i , j , k). Proof


Balanced Two-Way ANOVA Least-Squares Estimation

Fitted Values and Residuals

Form of fitted values depends on fit model:Additive: µjk = yj· + y·k − y··Interaction: µjk = yjk

Residuals have the form

eijk = yijk − µjk

where form of µjk depends on fit model (additive versus interaction).


Balanced Two-Way ANOVA Basic Inference

ANOVA Sums-of-Squares

In balanced two-way ANOVA model with interaction:SST =

∑bk=1

∑aj=1∑n∗

i=1(yijk − y··)2 df = abn∗ − 1

SSR = n∗∑b

k=1∑a

j=1(yjk − y··)2 df = ab − 1

SSE =∑b

k=1∑a

j=1∑n∗

i=1(yijk − yjk )2 df = abn∗ − ab

In balanced two-way ANOVA model with no interaction:SST =

∑bk=1

∑aj=1∑n∗

i=1(yijk − y··)2 df = abn∗ − 1

SSR = n∗∑b

k=1∑a

j=1([yj· + y·k − y··]− y··)2 df = a + b − 2

SSE =∑b

k=1∑a

j=1∑n∗

i=1(yijk − [yj· + y·k − y··])2

df = abn∗ − (a + b − 1)



Partitioning the Variance

From MLR notes we know that SST = SSR + SSE .

If njk = n∗∀j , k can partition SSR = SSA + SSB + SSAB whereSSA = bn∗

∑aj=1 α

2j df = a− 1

SSB = an∗∑b

k=1 β2k df = b − 1

SSAB = n∗∑b

k=1∑a

j=1(αβ)2jk df = (a− 1)(b − 1)

Implies that SSR = SSA + SSB for additive model (if njk = n∗∀j , k ).

Proof



Extended ANOVA Table and F Tests

We typically organize the SS information into an ANOVA table:Source SS df MS F p-valueSSR n∗

∑bk=1

∑aj=1(yjk − y··)2 ab − 1 MSR F∗ p∗

SSA bn∗∑a

j=1(yj· − y··)2 a− 1 MSA F∗a p∗aSSB an∗

∑bk=1(y·k − y··)2 b − 1 MSB F∗b p∗b

SSAB n∗∑b

k=1∑a

j=1(yjk − yj· − y·k + y··)2 (a− 1)(b − 1) MSAB F∗ab p∗abSSE

∑bk=1

∑aj=1∑n∗

i=1(yijk − yjk )2 ab(n∗ − 1) MSESST

∑bk=1

∑aj=1∑n∗

i=1(yijk − y··)2 abn∗ − 1

MSR = SSRab−1 , MSA = SSA

a−1 , MSB = SSBb−1 , MSAB = SSAB

(a−1)(b−1), MSE = SSE

ab(n∗−1),

F∗ = MSRMSE ∼ Fab−1,ab(n∗−1) and p∗ = P(Fab−1,ab(n∗−1) > F∗),

F∗a = MSAMSE ∼ Fa−1,ab(n∗−1) and p∗a = P(Fa−1,ab(n∗−1) > F∗a ),

F∗b = MSBMSE ∼ Fb−1,ab(n∗−1) and p∗b = P(Fb−1,ab(n∗−1) > F∗b ),

F∗ab = MSABMSE ∼ F(a−1)(b−1),ab(n∗−1) and p∗ab = P(F(a−1)(b−1),ab(n∗−1) > F∗ab),



ANOVA Table F Tests

F ∗ and p∗-value are testing H0 : αj = βk = (αβ)jk = 0 ∀j , k versusH1 : (∃j , k ∈ {1, . . . ,a} × {1, . . . ,b})(αj = βk = (αβ)jk = 0 is false)

Equivalent to H0 : µjk = µ ∀j , k versus H1 : not all µjk are equal

F ∗a statistic and p∗a-value are testing H0 : αj = 0 ∀j versusH1 : (∃j ∈ {1, . . . ,a})(αj 6= 0)

Testing main effect of first factor

F ∗b statistic and p∗b-value are testing H0 : βk = 0 ∀k versusH1 : (∃k ∈ {1, . . . ,b})(βk 6= 0)

Testing main effect of second factor

F ∗ab statistic and p∗ab-value are testing H0 : (αβ)jk = 0 ∀j , k versusH1 : (∃j , k ∈ {1, . . . ,a} × {1, . . . ,b})((αβ)jk 6= 0)

Testing interaction effectNathaniel E. Helwig (U of Minnesota) Factorial & Unbalanced Analysis of Variance Updated 04-Jan-2017 : Slide 14

Balanced Two-Way ANOVA Hypertension Example (Part 1)

Hypertension Example: Data Description

Hypertension example from Maxwell & Delany (2003).

Total of n = 72 subjects participate in hypertension experiment.Factor A: drug type (a = 3 levels: X, Y, Z)Factor B: diet type (b = 2 levels: yes, no)

Randomly assign njk = 12 subjects to each treatment cell:Note there are (ab) = (3)(2) = 6 treatment cellsObservations are independent within and between cells



Hypertension Example: Descriptive Statistics

Sum of blood pressure for each treatment cell (∑12

i=1 yijk ):Diet

Drug No (k = 1) Yes (k = 2) TotalX (j = 1) 2136 2052 4188Y (j = 2) 2424 2154 4578Z (j = 3) 2388 2130 4518

Total 6948 6336 13284

Sum-of-squares of blood pressure for each treatment cell (∑12

i=1 y2ijk ):

DietDrug No (k = 1) Yes (k = 2) Total

X (j = 1) 382368 352518 734886Y (j = 2) 491008 388898 879906Z (j = 3) 478238 380462 858700

Total 1351614 1121878 2473492Nathaniel E. Helwig (U of Minnesota) Factorial & Unbalanced Analysis of Variance Updated 04-Jan-2017 : Slide 16


Hypertension Example: OLS Estimation (by hand)

Least-squares estimates are cell means: µjk = yjk and

µ =1

abn∗

∑bk=1

∑aj=1∑n∗

i=1 yijk = y·· = 1328472 = 184.5

α1 =

(1

bn∗

∑bk=1

∑n∗i=1 yi1k

)− µ = y1· − y·· =

418824− 184.5 = −10

α2 =

(1

bn∗

∑bk=1

∑n∗i=1 yi2k

)− µ = y2· − y·· =

457824− 184.5 = 6.25

α3 =

(1

bn∗

∑bk=1

∑n∗i=1 yi3k

)− µ = y3· − y·· =

451824− 184.5 = 3.75

β1 =

(1

an∗

∑aj=1∑n∗

i=1 yij1

)− µ = y·1 − y·· =

694836− 184.5 = 8.5

β2 =

(1

an∗

∑aj=1∑n∗

i=1 yij2

)− µ = y·2 − y·· =

633636− 184.5 = −8.5



Hypertension Example: OLS Estimation (by hand)

Continuing from the previous slide. . .

(αβ)11 = y11 − y1· − y·1 + y·· =2136

12− 4188

24− 6948

36+ 184.5 = −5

(αβ)12 = y12 − y1· − y·2 + y·· =2052

12− 4188

24− 6336

36+ 184.5 = 5

(αβ)21 = y21 − y2· − y·1 + y·· =2424

12− 4578

24− 6948

36+ 184.5 = 2.75

(αβ)22 = y22 − y2· − y·2 + y·· =2154

12− 4578

24− 6336

36+ 184.5 = −2.75

(αβ)31 = y31 − y3· − y·1 + y·· =2388

12− 4518

24− 6948

36+ 184.5 = 2.25

(αβ)32 = y32 − y3· − y·2 + y·· =2130

12− 4518

24− 6336

36+ 184.5 = −2.25



Hypertension Example: Enter Data (in R)

> bp = scan("/Users/Nate/Desktop/hypertension.dat")Read 72 items> diet = factor(rep(rep(c("no","yes"),each=6),6))> drug = factor(rep(rep(c("X","Y","Z"),each=12),2))> biof = factor(rep(c("present","absent"),each=36))> hyper = data.frame(bp=bp, diet=diet, drug=drug, biof=biof)> hyper[1:20,]

bp diet drug biof1 170 no X present2 175 no X present3 165 no X present4 180 no X present5 160 no X present6 158 no X present7 161 yes X present8 173 yes X present9 157 yes X present10 152 yes X present11 181 yes X present12 190 yes X present13 186 no Y present14 194 no Y present15 201 no Y present16 215 no Y present17 219 no Y present18 209 no Y present19 164 yes Y present20 166 yes Y present



Hypertension Example: OLS Estimation (in R)Effect coding for drug and diet:> contrasts(hyper$drug) <- contr.sum(3)> contrasts(hyper$drug)

[,1] [,2]X 1 0Y 0 1Z -1 -1> contrasts(hyper$diet) <- contr.sum(2)> contrasts(hyper$diet)

[,1]no 1yes -1> mymod = lm(bp ~ drug * diet, data=hyper)> summary(mymod) # I deleted some output

Coefficients:Estimate Std. Error t value Pr(>|t|)

(Intercept) 184.500 1.642 112.355 < 2e-16 ***drug1 -10.000 2.322 -4.306 5.64e-05 ***drug2 6.250 2.322 2.691 0.00901 **diet1 8.500 1.642 5.176 2.30e-06 ***drug1:diet1 -5.000 2.322 -2.153 0.03498 *drug2:diet1 2.750 2.322 1.184 0.24059---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 13.93 on 66 degrees of freedomMultiple R-squared: 0.4329, Adjusted R-squared: 0.3899F-statistic: 10.07 on 5 and 66 DF, p-value: 3.385e-07



Hypertension Example: Sums-of-Squares (by hand 1)

Defining n =∑b

k=1∑a

j=1 njk , the relevant sums-of-squares are

SST =∑b

k=1∑a

j=1∑njk

i=1(yijk − y··)2 =∑b

k=1∑a

j=1∑njk

i=1 y2ijk −

1n

(∑bk=1

∑aj=1∑njk

i=1 yijk

)2

= 2473492−172

(13284)2 = 22594

SSE =∑b

k=1∑a

j=1∑njk

i=1(yijk − yjk )2 =∑b

k=1∑a

j=1∑njk

i=1 y2ijk −

∑bk=1

∑aj=1

(∑njki=1 yijk

)2

njk

= 2473492−[21362 + 20522 + 24242 + 21542 + 23882 + 21302

]/12 = 12814

SSR = SST − SSE = 22594− 12814 = 9780



Hypertension Example: Sums-of-Squares (by hand 2)The sums-of-squares for the main and interaction effects are given by

SSA = bn∗∑a

j=1(yj· − y··)2 = bn∗∑a

j=1 α2j

= (2)(12)[(−10)2 + 6.252 + 3.752

]= 3675

SSB = an∗∑b

k=1(y·k − y··)2 = an∗∑b

k=1 β2k

= (3)(12)[(−8.5)2 + 8.52

]= 5202

SSAB = n∗∑b

k=1∑a

j=1(yjk − yj· − y·k + y··)2 = n∗∑b

k=1∑a

j=1(αβ)2jk

= 12[(−5)2 + 52 + 2.752 + (−2.75)2 + 2.252 + (−2.25)2

]= 903

and since njk = n∗ = 12 ∀j , k , we have

SSR = SSA + SSB + SSAB9780 = 3675 + 5202 + 903



Hypertension Example: ANOVA Table (by hand)

Putting things together in ANOVA table:Source SS df MS F p-valueSSR 9780 5 1956.0 10.07 < .0001

SSA 3675 2 1837.5 9.46 0.0002SSB 5202 1 5202.0 26.79 < .0001SSAB 903 2 451.5 2.33 0.1057

SSE 12814 66 194.2SST 22594 71



Hypertension Example: ANOVA Table (in R)

> anova(mymod)Analysis of Variance Table

Response: bpDf Sum Sq Mean Sq F value Pr(>F)

drug 2 3675 1837.5 9.4643 0.0002433 ***diet 1 5202 5202.0 26.7935 2.305e-06 ***drug:diet 2 903 451.5 2.3255 0.1056925Residuals 66 12814 194.2---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


Balanced Two-Way ANOVA Multiple Comparisons

Multiple Comparisons Overview

Still have multiple comparison problem:Overall test is not very informativeCan examine effect estimates for group differencesNeed follow-up tests to examine linear combinations of means

Still can use the same tools as before:BonferroniTukey (Tukey-Kramer)Scheffé



Two-Way ANOVA Linear Combinations

Assuming interaction model, we now have

L =∑b

k=1∑a

j=1 cjk yjk and V (L) = σ2∑bk=1

∑aj=1 c2

jk/njk

where cjk are the coefficients and σ2 is the MSE.

Assuming the additive model, we have

La =∑a

j=1 cj yj· and V (La) = σ2∑aj=1 c2

j /nj·

Lb =∑b

k=1 ck y·k and V (Lb) = σ2∑bk=1 c2

k/n·k

where cj and ck are main effect coefficients, σ2 is the MSE, andnj· =

∑bk=1 njk and n·k =

∑aj=1 njk are the marginal sample sizes.



Two-Way Multiple Comparisons in Practice

For interaction model, you follow-up on µjk = yjk

Bonferroni for any f tests (independent or not)Tukey (Tukey-Kramer) for all pairwise comparisonsScheffé for all possible contrasts

For additive model, you follow-up on µj = yj· and µk = y·kBonferroni for any f tests (independent or not)Tukey (Tukey-Kramer) for all pairwise comparisonsScheffé for all possible contrasts

For additive model, Tukey and Scheffé control FWER for each maineffect family separately.

Use Bonferroni in combination with Tukey/Scheffé to controlFWER for both families simultaneously



Hypertension Example: Interaction (by hand)

All ab(ab − 1)/2 = 15 possible pairwise comparisons of µjk :

L = yjk − yj ′k ′ and V (L) = 194.2(2/12) = 32.36667

and we know that√

2(L)√V (L)∼ qab,abn∗−ab, so 100(1− α)% CI is given by

L± 1√2

q(α)ab,abn∗−ab

√V (L)

where q(α)ab,abn∗−ab is critical value from studentized range.

For example, 95% CI for µ21 − µ11 is given by:

(µ21 − µ11)± 1√2

q(.05)6,66

√V (L)(

242412− 2136

12

)± 1√

2(4.150851)

√32.36667 = [7.303829; 40.69617]



Hypertension Example: Interaction (in R)

All ab(ab − 1)/2 = 15 possible pairwise comparisons of µjk :> mymod = aov(bp ~ drug * diet, data=hyper)> TukeyHSD(mymod, "drug:diet")

Tukey multiple comparisons of means95% family-wise confidence level

Fit: aov(formula = bp ~ drug * diet, data = hyper)

$‘drug:diet‘diff lwr upr p adj

Y:no-X:no 24.0 7.303829 40.696171 0.0010415Z:no-X:no 21.0 4.303829 37.696171 0.0058124X:yes-X:no -7.0 -23.696171 9.696171 0.8203137Y:yes-X:no 1.5 -15.196171 18.196171 0.9998189Z:yes-X:no -0.5 -17.196171 16.196171 0.9999992Z:no-Y:no -3.0 -19.696171 13.696171 0.9948741X:yes-Y:no -31.0 -47.696171 -14.303829 0.0000117Y:yes-Y:no -22.5 -39.196171 -5.803829 0.0025081Z:yes-Y:no -24.5 -41.196171 -7.803829 0.0007710X:yes-Z:no -28.0 -44.696171 -11.303829 0.0000856Y:yes-Z:no -19.5 -36.196171 -2.803829 0.0128988Z:yes-Z:no -21.5 -38.196171 -4.803829 0.0044123Y:yes-X:yes 8.5 -8.196171 25.196171 0.6690751Z:yes-X:yes 6.5 -10.196171 23.196171 0.8616371Z:yes-Y:yes -2.0 -18.696171 14.696171 0.9992610



Hypertension Example: Additive (by hand A part 1)

All a(a− 1)/2 = 3 possible pairwise comparisons of µj :

Y− X : La1 =4578

24− 4188

24= 16.25

Z− X : La2 =4518

24− 4188

24= 13.75

Z− Y : La3 =4518

24− 4578

24= −2.5

and the variance is given by

V (Laj ) = σ2∑aj=1 c2

j /nj· = (201.7206)(2/24) = 16.81005

where σ2 = SSE+SSABabn∗−(a+b−1) = 12814+903

68 = 201.7206



Hypertension Example: Additive (by hand A part 2)

Note√

2(Laj )√V (Laj )

∼ qa,abn∗−(a+b−1), so 100(1− α)% CI is given by

Laj ±1√2

q(α)a,abn∗−(a+b−1)

√V (Laj )

where q(α)a,abn∗−(a+b−1) is critical value from studentized range.

The 95% CI for all three pairwise comparisons is given by

La1 ±1√2

q(.05)3,68

√V (La1 ) = 16.25± 1√

2(3.388576)

√16.81005 = [6.426037; 26.07396]

La2 ±1√2

q(.05)3,68

√V (La2 ) = 13.75± 1√

2(3.388576)

√16.81005 = [3.926037; 23.57396]

La3 ±1√2

q(.05)3,68

√V (La3 ) = −2.5± 1√

2(3.388576)

√16.81005 = [−12.32396; 7.323963]



Hypertension Example: Additive (by hand B part 1)

All b(b − 1)/2 = 1 possible pairwise comparison of µk :

yes− no : Lb =6336

36− 6948

36= −17

and the variance is given by

V (Lb) = σ2∑bk=1 c2

k/n·k = (201.7206)(2/36) = 11.2067

where σ2 = SSE+SSABabn∗−(a+b−1) = 12814+903

68 = 201.7206



Hypertension Example: Additive (by hand B part 2)

Note√

2(Lb)√V (Lb)

∼ qb,abn∗−(a+b−1), so 100(1− α)% CI is given by

Lb ±1√2

q(α)b,abn∗−(a+b−1)

√V (Lb)

where q(α)b,abn∗−(a+b−1) is critical value from studentized range.

The 95% CI for pairwise comparison is given by

Lb ±1√2

q(.05)2,68

√V (Lb) = −17± 1√

2(2.822019)

√11.2067

= [−23.68011; −10.31989]



Hypertension Example: Additive (in R)All a(a− 1)/2 = 3 possible pairwise comparisons of µj :> mymod = aov(bp ~ drug + diet, data=hyper)> TukeyHSD(mymod, "drug")


Fit: aov(formula = bp ~ drug + diet, data = hyper)

$drugdiff lwr upr p adj

Y-X 16.25 6.426037 26.073963 0.0005220Z-X 13.75 3.926037 23.573963 0.0036941Z-Y -2.50 -12.323963 7.323963 0.8152941

All b(b − 1)/2 = 1 possible pairwise comparison of µk :> mymod = aov(bp ~ drug + diet, data=hyper)> TukeyHSD(mymod,"diet")


Fit: aov(formula = bp ~ drug + diet, data = hyper)

$dietdiff lwr upr p adj

yes-no -17 -23.68011 -10.31989 3.2e-06


Balanced Three-Way ANOVA

Balanced Three-Way ANOVA


Balanced Three-Way ANOVA Model Form and Estimation

Three-Way ANOVA Model (cell means form)

The Three-Way Analysis of Variance (ANOVA) model has the form

yijkl = µjkl + eijkl

for i ∈ {1, . . . ,njkl}, j ∈ {1, . . . ,a}, k ∈ {1, . . . ,b}, l ∈ {1, . . . , c}, whereyijkl ∈ R is response for i-th subject in factor cell (j , k , l)µjkl ∈ R is population mean for factor cell (j , k , l)

eijkliid∼ N(0, σ2) is a Gaussian error term

njkl is number of subjects in cell (j , k , l)(note: njkl = n∗∀j , k , l in balanced three-way ANOVA)(a,b, c) is number of factor levels for Factors (A,B,C)

Implies that yijklind∼ N(µjkl , σ

2).



OLS Estimation (cell means form)

Similar to balanced two-way ANOVA, we want to minimize

c∑l=1

b∑k=1

a∑j=1

njkl∑i=1

(yijkl − µjkl)2

which is equivalent to minimizing∑njkl

i=1(yijkl − µjkl)2 for all j , k , l

Taking the derivative of SSEjkl =∑njkl

i=1(yijkl − µjkl)2, we see that

dSSEjkl

dµjkl= −2

njkl∑i=1

yijkl + 2njklµjkl

and setting to zero and solving gives µjkl = 1njkl

∑njkli=1 yijkl = yjkl



Three-Way ANOVA Model (effect coding)

The three-way ANOVA with all interactions assumes that

µjkl = µ+ αj + βk + γl + (αβ)jk + (αγ)jl + (βγ)kl + (αβγ)jkl

for j ∈ {1, . . . ,a}, k ∈ {1, . . . ,b}, and l ∈ {1, . . . , c} whereµ is overall population mean

αj is main effect of first factor such that∑a

j=1 αj = 0

βk is main effect of second factor such that∑b

k=1 βk = 0

γl is main effect of third factor such that∑c

l=1 γl = 0

(αβ)jk is A ∗ B interaction effect such that∑a

j=1(αβ)jk = 0 ∀k and∑b

k=1(αβ)jk = 0 ∀j

(αγ)jl is A ∗ C interaction effect such that∑a

j=1(αγ)jl = 0 ∀l and∑c

l=1(αγ)jl = 0 ∀j

(βγ)kl is B ∗ C interaction effect such that∑b

k=1(βγ)kl = 0 ∀l and∑c

l=1(βγ)kl = 0 ∀k(αβγ)jkl is A ∗ B ∗ C interaction effect such that

∑aj=1(αβγ)jkl = 0 ∀k , l and∑b

k=1(αβγ)jkl = 0 ∀j, l and∑c

l=1(αβγ)jkl = 0 ∀j, k



OLS Estimation (effect coding)

The OLS estimates of the various effects are given by

µ = y···αj = yj·· − y···βk = y·k · − y···γl = y··l − y···

(αβ)jk = yjk · − yj·· − y·k · + y···(αγ)jl = yj·l − yj·· − y··l + y···(βγ)kl = y·kl − y·k · − y··l + y···

( ˆαβγ)jkl = yjkl − [µ+ αj + βk + γl + (αβ)jk + (αγ)jl + (βγ)kl ]



Fitted Values and Residuals

Form of fitted values depends on fit model:Additive: µjkl = µ+ αj + βk + γl

All 2-way Int: µjkl = µ+ αj + βk + γl + (αβ)jk + (αγ)jl + (βγ)kl

3-way Int: µjkl = yjkl

Residuals have the form

eijkl = yijkl − µjkl

where form of µjkl depends on fit model (additive versus interaction).


Balanced Three-Way ANOVA Basic Inference

ANOVA Sums-of-Squares

Defining N = abcn∗, the three-way ANOVA sums-of-squares are:SST =

∑cl=1∑b

k=1∑a

j=1∑n∗

i=1(yijkl − y···)2 df = N − 1SSR = n∗

∑cl=1∑b

k=1∑a

j=1(yjkl − y···)2 df = abc − 1SSE =

∑cl=1∑b

k=1∑a

j=1∑n∗

i=1(yijkl − yjkl)2 df = N − abc

SSR = SSA + SSB + SSC + SSAB + SSAC + SSBC + SSABCSSA = bcn∗

∑aj=1 α

2j df = a− 1

SSB = acn∗∑b

k=1 β2k df = b − 1

SSC = abn∗∑c

l=1 γ2l df = c − 1

SSAB = cn∗∑b

k=1∑a

j=1(αβ)2jk df = (a− 1)(b − 1)

SSAC = bn∗∑c

l=1∑a

j=1(αγ)2jl df = (a− 1)(c − 1)

SSBC = an∗∑c

l=1∑b

k=1(βγ)2kl df = (b − 1)(c − 1)

SSABC = n∗∑c

l=1∑b

k=1∑a

j=1( ˆαβγ)2jkl

df = (a− 1)(b − 1)(c − 1)



Memory Example: Data Description (revisited)

Hypertension example from Maxwell & Delany (2003).

Total of N = 72 subjects participate in hypertension experiment.Factor A: drug type (a = 3 levels: X, Y, Z)Factor B: diet type (b = 2 levels: yes, no)Factor C: biof type (c = 2 levels: present, absent)

Randomly assign njkl = 6 subjects to each treatment cell:Note there are (abc) = (3)(2)(2) = 12 treatment cellsObservations are independent within and between cells



Hypertension Example: Look at Data

> bp = scan("/Users/Nate/Desktop/hypertension.dat")Read 72 items> diet = factor(rep(rep(c("no","yes"),each=6),6))> drug = factor(rep(rep(c("X","Y","Z"),each=12),2))> biof = factor(rep(c("present","absent"),each=36))> hyper = data.frame(bp=bp, diet=diet, drug=drug, biof=biof)> contrasts(hyper$drug) <- contr.sum(3)> contrasts(hyper$diet) <- contr.sum(2)> contrasts(hyper$biof) <- contr.sum(2)> hyper[1:15,]

bp diet drug biof1 170 no X present2 175 no X present3 165 no X present4 180 no X present5 160 no X present6 158 no X present7 161 yes X present8 173 yes X present9 157 yes X present10 152 yes X present11 181 yes X present12 190 yes X present13 186 no Y present14 194 no Y present15 201 no Y present



Hypertension Example: All Interactions

> mymod = lm(bp ~ drug * diet * biof, data=hyper)> anova(mymod)Analysis of Variance Table


drug 2 3675 1837.5 11.7287 5.019e-05 ***diet 1 5202 5202.0 33.2043 3.053e-07 ***biof 1 2048 2048.0 13.0723 0.0006151 ***drug:diet 2 903 451.5 2.8819 0.0638153 .drug:biof 2 259 129.5 0.8266 0.4424565diet:biof 1 32 32.0 0.2043 0.6529374drug:diet:biof 2 1075 537.5 3.4309 0.0388342 *Residuals 60 9400 156.7---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1



Hypertension Example: All 2-Way Interactions

> mymod = lm(bp ~ drug * diet + drug * biof + diet * biof, data=hyper)> anova(mymod)Analysis of Variance Table


drug 2 3675 1837.5 10.8759 8.940e-05 ***diet 1 5202 5202.0 30.7899 6.345e-07 ***biof 1 2048 2048.0 12.1218 0.000919 ***drug:diet 2 903 451.5 2.6724 0.077043 .drug:biof 2 259 129.5 0.7665 0.468992diet:biof 1 32 32.0 0.1894 0.664925Residuals 62 10475 169.0---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1



Hypertension Example: Additive Model

> mymod = lm(bp ~ drug + diet + biof, data=hyper)> anova(mymod)Analysis of Variance Table


drug 2 3675 1837.5 10.550 0.0001039 ***diet 1 5202 5202.0 29.868 7.346e-07 ***biof 1 2048 2048.0 11.759 0.0010403 **Residuals 67 11669 174.2---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1



Hypertension Example: Interaction Plot

If you choose the three-way interaction model, you could visualize theinteraction using an interaction plot.

Biofeedback Absent

Drug

Mea

n B

P

Diet NoDiet Yes

X Y Z

170

180

190

200

210

Biofeedback Present

Drug

Mea

n B

P

Diet NoDiet Yes

X Y Z

170

180

190

200

210



Hypertension Example: Interaction Plot (R code)

yhat=tapply(hyper$bp,list(hyper$drug,hyper$diet,hyper$biof),mean)par(mfrow=c(1,2))mytitles=c("Biofeedback Absent","Biofeedback Present")for(k in 1:2){plot(1:3,yhat[,1,k],ylim=c(165,215),xlab="Drug",

ylab="Mean BP",main=mytitles[k],axes=FALSE,type="l")lines(1:3,yhat[,2,k],lty=2)legend("topleft",c("Diet No","Diet Yes"),lty=1:2,bty="n")axis(1,at=1:3,labels=c("X","Y","Z"))axis(2)

}



Hypertension Example: Multiple Comparisons

Assuming we chose the additive model, we would perform follow-uptests on the marginal means.

Factor A: µaj = µ+ αj = yj··

Factor B: µbk = µ+ βk = y·k ·Factor C: µcl = µ+ γl = y··l

If we chose three-way interaction model, we would perform follow-uptests on the individual cell means.

µjkl = µ+ αj + βk + γl + (αβ)jk + (αγ)jl + (βγ)kl + ( ˆαβγ)jkl

= yjkl



Hypertension Example: Multiple Comparisons

> mymod = aov(bp ~ drug + diet + biof, data=hyper)

> TukeyHSD(mymod, "drug") # I deleted some outputTukey multiple comparisons of means

95% family-wise confidence level

$drugdiff lwr upr p adj

Y-X 16.25 7.118642 25.381358 0.0001874Z-X 13.75 4.618642 22.881358 0.0016810Z-Y -2.50 -11.631358 6.631358 0.7894946

> TukeyHSD(mymod, "diet") # I deleted some outputTukey multiple comparisons of means


$dietdiff lwr upr p adj

yes-no -17 -23.20877 -10.79123 7e-07

> TukeyHSD(mymod, "biof") # I deleted some outputTukey multiple comparisons of means


$biofdiff lwr upr p adj

present-absent -10.66667 -16.87544 -4.457897 0.0010403


Unbalanced ANOVA Models

Unbalanced ANOVA Models


Unbalanced ANOVA Models Overview of Problem

Unbalanced ANOVA: Model Form

Unbalanced ANOVA has same model form as balanced, but unequalsample sizes in each cell.

1-way: nj 6= nj ′ for some j , j ′

2-way: njk 6= nj ′k ′ for some (jk), (j ′k ′)3-way: njkl 6= nj ′k ′l ′ for some (jkl), (j ′k ′l ′)

In the previous slides, we assumed njk = n∗∀j , k (two-way ANOVA) ornjkl = n∗∀j , k , l (three-way ANOVA), which made life easy.

Effects were orthogonal in balanced designParameter estimates had simple relation to cell/marginal means



Unbalanced ANOVA: Implications

Main consequence for two-way (and higher-way) unbalanced design:Non-orthogonal SS (e.g., SSR 6= SSA + SSB + SSAB)Design is less efficient (larger variances of parameter estimates)

Unbalanced design also affects our estimation and follow-up testsParameter estimates require matrix inversion: b = (X′X)−1X′yNeed to do follow-up tests on least-squares means



Unbalanced ANOVA: Testing Effects

Because of non-orthogonality, cannot test effects using F = MS?MSE .

Instead we use the General Linear Model (GLM) F test statistic:

F =SSER − SSEF

dfR − dfF÷ SSEF

dfF∼ F(dfR−dfF ,dfF )

whereSSER is sum-of-squares error for reduced modelSSEF is sum-of-squares error for full modeldfR is error degrees of freedom for reduced modeldfF is error degrees of freedom for full model



Unbalanced ANOVA: Testing Example

Consider two-way ANOVA and all 7 possible models

yijk = µ+ αj + βk + (αβ)jk + eijk (1)yijk = µ+ αj + βk + eijk (2)yijk = µ+ αj + (αβ)jk + eijk (3)yijk = µ+ βk + (αβ)jk + eijk (4)yijk = µ+ αj + eijk (5)yijk = µ+ βk + eijk (6)yijk = µ+ eijk (7)



Unbalanced ANOVA: Testing Example (continued)

To test effect, use F test comparing full and reduced models.

To test each effect there are multiple choices we could use for full andreduced models:

A: F=1 and R=4 or F=2 and R=6 or F=5 and R=7B: F=1 and R=3 or F=2 and R=5 or F=6 and R=7AB: F=1 and R=2 or F=3 and R=5 or F=4 and R=6


Unbalanced ANOVA Models Types of Sums-of-Squares

Types of Sum-of-SquaresType I SS

Amount of additional variation explained by the model when a term is added to the model(aka sequential sum-of-squares).In two-way ANOVA, type I SS would compare:(a) Main Effect A: F=5 and R=7(b) Main Effect B: F=2 and R=5(c) Interaction Effect: F=1 and R=2

Type II SSAmount of variation a term adds to the model when all other terms are included exceptterms that “contain” the effect being tested (e.g., (αβ)jk contains αj and βk ).In two-way ANOVA, type II SS would compare:(a) Main Effect A: F=2 and R=6(b) Main Effect B: F=2 and R=5(c) Interaction Effect: F=1 and R=2

Type III SSAmount of variation a term adds to the model when all other terms are included, which issometimes called partial sum-of-squares.In two-way ANOVA, type III SS would compare:(a) Main Effect A: F=1 and R=4(b) Main Effect B: F=1 and R=3(c) Interaction Effect: F=1 and R=2


Unbalanced ANOVA Models Types of Sums-of-Squares

Types of Sum-of-Squares (in R)

When fitting multi-way ANOVAs, anova function gives Type I SS.

Order matters in unbalanced design!bp = drug + diet produces different Type I SS tests thanbp = diet + drug if design is unbalanced

Use Anova function in car package for Type II and Type III SS.Function performs Type II SS tests by defaultUse type=3 option for Type III SS tests


Unbalanced ANOVA Models Hypertension Example (Part 4)

Hypertension Example: Type I

> mymod = lm(bp ~ drug * diet * biof, data=hyper[1:71,])> anova(mymod)Analysis of Variance Table


drug 2 3733.6 1866.8 11.7306 5.138e-05 ***diet 1 5113.3 5113.3 32.1311 4.558e-07 ***biof 1 2087.2 2087.2 13.1154 0.0006101 ***drug:diet 2 879.5 439.8 2.7633 0.0712569 .drug:biof 2 280.5 140.3 0.8813 0.4196123diet:biof 1 24.2 24.2 0.1522 0.6978384drug:diet:biof 2 1055.8 527.9 3.3172 0.0431275 *Residuals 59 9389.2 159.1---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1



Hypertension Example: Type II

> library(car)> Anova(mymod, type=2)Anova Table (Type II tests)

Response: bpSum Sq Df F value Pr(>F)

drug 3704.1 2 11.6378 5.491e-05 ***diet 4975.9 1 31.2676 6.085e-07 ***biof 2061.8 1 12.9561 0.0006541 ***drug:diet 872.5 2 2.7413 0.0727049 .drug:biof 277.7 2 0.8726 0.4231893diet:biof 24.2 1 0.1522 0.6978384drug:diet:biof 1055.8 2 3.3172 0.0431275 *Residuals 9389.2 59---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1



Hypertension Example: Type III

> library(car)> Anova(mymod, type=3)Anova Table (Type III tests)

Response: bpSum Sq Df F value Pr(>F)

(Intercept) 2412026 1 15156.7271 < 2.2e-16 ***drug 3685 2 11.5784 5.730e-05 ***diet 5057 1 31.7754 5.132e-07 ***biof 2052 1 12.8967 0.0006713 ***drug:diet 882 2 2.7705 0.0707910 .drug:biof 268 2 0.8434 0.4353639diet:biof 27 1 0.1692 0.6822873drug:diet:biof 1056 2 3.3172 0.0431275 *Residuals 9389 59---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


Appendix

Appendix


Appendix

Ordinary Least-Squares (proof for µ)

Expanding the first summation produces

SSE =b∑

k=1

a∑j=1

[n∗∑i=1

y2ijk − 2(µ+ αj + βk + (αβ)jk )

n∗∑i=1

yijk + n∗(µ+ αj + βk + (αβ)jk )2

]

Taking the derivative with respect to µ we have

dSSEdµ

=∑b

k=1∑a

j=1[−2∑n∗

i=1 yijk + 2n∗µ+ 2n∗(αj + βk + (αβ)jk )]

= −2(∑b

k=1∑a

j=1∑n∗

i=1 yijk

)+ 2abn∗µ

and setting to zero and solving for µ givesµ = 1

abn∗

∑bk=1

∑aj=1∑n∗

i=1 yijk = y··


Appendix

Ordinary Least-Squares (proof for αj)

Taking the derivative with respect to αj we have

dSSEdαj

=∑b

k=1[−2∑n∗

i=1 yijk + 2n∗αj + 2n∗(µ+ βk + (αβ)jk )]

= −2(∑b

k=1∑n∗

i=1 yijk

)+ 2bn∗αj + 2bn∗µ

and setting to zero, using µ for µ, and solving for αj givesαj = 1

bn∗ (∑b

k=1∑n∗

i=1 yijk )− µ = yj· − y··


Appendix

Ordinary Least-Squares (proof for βk )

Taking the derivative with respect to βk we have

dSSEdβk

=∑a

j=1[−2∑n∗

i=1 yijk + 2n∗βk + 2n∗(µ+ αj + (αβ)jk )]

= −2(∑a

j=1∑n∗

i=1 yijk

)+ 2an∗βk + 2an∗µ

and setting to zero, using µ for µ, and solving for βk givesβk = 1

an∗ (∑a

j=1∑n∗

i=1 yijk )− µ = y·k − y··


Appendix

Ordinary Least-Squares (proof for (αβ)jk )

Taking the derivative with respect to (αβ)jk we have

dSSEd(αβ)jk

= −2n∗∑

i=1

yijk + 2n∗(αβ)jk + 2n∗(µ+ αj + βk )

and setting to zero, using (µ, αj , βk ) for (µ, αj , βk ), and solving for(αβ)jk gives (αβ)jk = 1

n∗ (∑n∗

i=1 yijk )− µ− αj − βk = yjk − yj· − y·k + y··

Return


Appendix

Partitioning the Variance (proof part 1)To prove SSR = SSA + SSB + SSAB when njk = n∗∀j , k , note that

yijk − y·· = (yijk − yjk ) + (yjk − [yj· + y·k − y··]) + (yj· − y··) + (y·k − y··)

Now if we square both sides we have(yijk − y··)2 = (yijk − yjk )2 + (yjk − [yj· + y·k − y··])2 + (yj· − y··)2 + (y·k − y··)2

+ 2(yijk − yjk ) {(yjk − [yj· + y·k − y··]) + (yj· − y··) + (y·k − y··)}+ 2(yjk − [yj· + y·k − y··]) [(yj· − y··) + (y·k − y··)]

+ 2(yj· − y··)(y·k − y··)

Now if we apply the triple summation we have SST

SST =b∑

k=1

a∑j=1

njk∑i=1

(yijk − y··)2


Appendix

Partitioning the Variance (proof part 2)First, note that we have

SSE =∑b

k=1

∑aj=1

∑njki=1(yijk − yjk )2

SSAB =∑b

k=1

∑aj=1

∑njki=1(yjk − [yj· + y·k − y··])2

SSA =∑b

k=1

∑aj=1

∑njki=1(yj· − y··)2

SSB =∑b

k=1

∑aj=1

∑njki=1(y·k − y··)2

so we need to prove that the crossproduct terms are orthogonal.

To prove that the first crossproduct term sums to zero, define(αβ)jk = (yjk − [yj· + y·k − y··]) + (yj· − y··) + (y·k − y··) and note that∑b

k=1∑a

j=1∑njk

i=1 2(yijk − yjk )(αβ)jk = 2∑b

k=1∑a

j=1(αβ)jk∑njk

i=1(yijk − yjk )

= 2∑b

k=1∑a

j=1(αβ)jk (0) = 0

because we are summing mean-centered variable.Nathaniel E. Helwig (U of Minnesota) Factorial & Unbalanced Analysis of Variance Updated 04-Jan-2017 : Slide 68

Appendix

Partitioning the Variance (proof part 3)

To prove that the second crossproduct term sums to zero, note that(αβ)jk = (yjk − [yj· + y·k − y··]), αj = (yj· − y··), and βk = (y·k − y··), so∑b

k=1∑a

j=1∑njk

i=1 2(αβ)jk (αj + βk ) = 2∑b

k=1∑a

j=1 njk (αβ)jk (αj + βk )

Now assuming that njk = n∗∀j , k∑bk=1

∑aj=1 njk (αβ)jk αj = n∗

∑aj=1 αj

(∑bk=1(αβ)jk

)= n∗

∑aj=1 αj(0) = 0∑b

k=1∑a

j=1 njk (αβ)jk βk = n∗∑b

k=1 βk

(∑aj=1(αβ)jk

)= n∗

∑bk=1 βk (0) = 0


Appendix

Partitioning the Variance (proof part 4)

To prove that the third crossproduct term sums to zero, note that∑bk=1

∑aj=1∑njk

i=1 2(yj· − y··)(y·k − y··) = 2∑b

k=1∑a

j=1 njk αj βk

and if njk = n∗∀j , k we have that

2∑b

k=1∑a

j=1 njk αj βk = 2n∗∑b

k=1∑a

j=1 αj βk

= 2n∗∑b

k=1 βk

(∑aj=1 αj

)= 2n∗

∑bk=1 βk (0) = 0

which completes the proof.

Return


nathaniel e. helwig - umn statisticsusers.stat.umn.edu/~helwig/notes/aov2-notes.pdf ·...

Documents